I have a folder that contains versions of my application, each time I upload a new version a new sub-folder is created for it, the sub-folder name is the current timestamp, here
ls -dt1 /path/to/folder/*/ | sed '11,$p' | rm -r
this assumes those are the only directories and no others are present in the working directory.
ls -dt1 will normally only print the newest directory however the /*/ will
only match directories and print their full paths the 1 ensures one
line per match/listing t sorts time with newest at the top.
sed takes the 11th line on down to the bottom and prints only those lines, which are then passed to rm.
You can use xargs, but for testing you may wish to remove | rm -r to see if the directories are listed properly first.
Your directory names are sorted in chronological order, which makes this easy. The list of directories in chronological order is just *, or [0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9] to be more precise. So you want to delete all but the last 10 of them.
set [0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]/
while [ $# -gt 10 ]; do
rm -rf "$1"
shift
fi
(While there are more than 10 directories left, delete the oldest one.)
ls -lt | grep ^d | sed -e '1,10d' | awk '{sub(/.* /, ""); print }' | xargs rm -rf
Explanation:
use awk to extract the file names from the remaining 'ls -l' output
remove the files
EDIT:
find . -maxdepth 1 -type d ! -name \\.| sort | tac | sed -e '1,10d' | xargs rm -rf
There you go. (edited)
ls -dt */ | tail -n +11 | xargs rm -rf
First list directories recently modified then take all of them except first 10, then send them to rm -rf.
If the directories' names contain the date one can delete all but the last 10 directories with the default alphabetical sort
ls -d */ | head -n -10 | xargs rm -rf